I am not entirely sure what you are asking for question 2, but let me take a crack at (1). At verious points I will be sloppy about distiguishing between a coset and its chosen representative, but it shouldn't cause any confusion.

(1) We have $B(F)\backslash G(F) = \cup_{w\in W}B\backslash BwU$. To get the coset reps, we need to see when we can have $$bwu=b'w'u'\quad \text{or, more simply} \quad wu=bw'.$$

Since we know that the Bruhat decomposition is a disjoint union, for this to occur we must have $w'=w$, reducing the identity to $wuw^{-1}=b$. Note that this forces $b\in U(F)$, since $b\in B(F)$ and $wuw^{-1}\in G(F)$ is unipotent.

Thus, the elements of $U(F)$ such that $B(F)wu = B(F)w$ is precisely $(U(F)\cap wU(F)w^{-1})$. Hence, to get coset representatives for the Bruhat cell corresponding to $w\in W$, we need to select representatives of the quotient $$(U(F)\cap wU(F)w^{-1})\backslash U(F).$$

For (2), it seems that all they are doing is the standard change of variables $u\mapsto u(u')^{-1}$ with $u$ a representative in the quotient $$(U(A)\cap wU(A)w^{-1})\backslash U(A),$$ and $u'$ a representative of the quotient $$(U(F)\cap wU(F)w^{-1})\backslash (U(A)\cap wU(A)w^{-1}).$$

The reason $u'$ does not appear in the argument for $f$ is that $u'\in (U(A)\cap wU(A)w^{-1})$ allows you to push it past $w^{-1}$ and use the invariance properties of $f$ as an element of the parabolocally induced representation to get rid of it.